Work/Energy problem at an angle

  • Thread starter Thread starter Heather_
  • Start date Start date
  • Tags Tags
    Angle
Click For Summary
SUMMARY

The discussion centers on a physics problem involving work and energy as a block of ice slides down a frictionless ramp at an angle of θ = 50°. The worker applies a pulling force of Fr = 60 N, resulting in a kinetic energy increase of 85 J over a distance of d = 0.65 m. The correct calculation reveals that the gravitational force in the X direction is Fgx = 263.44 N, leading to a total work done of W = 110.06 J. The error identified was the unnecessary use of the cosine factor in the work equation, as both forces are parallel to the ramp.

PREREQUISITES
  • Understanding of work-energy principles
  • Familiarity with vector components of forces
  • Knowledge of gravitational force calculations
  • Proficiency in using the work equation W = F * d
NEXT STEPS
  • Review the concept of work done by forces on inclined planes
  • Study the implications of frictionless surfaces in physics problems
  • Learn about the relationship between work and kinetic energy changes
  • Explore vector decomposition of forces in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to work and energy on inclined planes.

Heather_
Messages
4
Reaction score
0
1. Homework Statement

In the figure below, a block of ice slides down a frictionless ramp at angle θ = 50°, while an ice worker pulls up the ramp (via a rope) with a force of magnitude Fr = 60 N. As the block slides through distance d = 0.65 m along the ramp, its kinetic energy increases by 85 J. How much greater would its kinetic energy have been if the rope had not been attached to the block?

2. Homework Equations
W=ΔEk
W=F*d*cosθ

3. The Attempt at a Solution

First, I solved for the gravitational force in the X direction: (Fgx-60N)*.65m*cos50=85J
Fgx = 263.44 N
Then, I plugged it back into the work equation to see how much work would be done then: 263.44N*.65m*cos50=W
W=110.06J=ΔEk
110.6J-85J=25.6J
Yet, this is not the right answer. What am I doing wrong?
 
Physics news on Phys.org
Heather_ said:
First, I solved for the gravitational force in the X direction: (Fgx-60N)*.65m*cos50=85J
The x-component of gravity and the 60 N force are already parallel to the ramp, so no need for the cos50 factor.
 
Wow. That was a silly mistake. Thank you! :)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 56 ·
2
Replies
56
Views
5K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K